I would like to have some explanation for the following statement

Let $K$ be an algebraically closed field of characteristic $p>0$, and $K((t))$, the field of Laurent series with coefficients in $K$. The Galois group of the polynomial $X^{p^n}-X=t^{-1}$ is isomorphic to the additive group of $F_{p^n}$, i.e. to $(\mathbb{Z}/p\mathbb{Z})^n$.

Another question:

Are there some extensions with Galois group isomorphic to $(\mathbb{Z}/p^n\mathbb{Z})$ with $n>1$

  • $\begingroup$ It is very rude to just change the problem statement after the problem has been answered. $\endgroup$ – Dedalus Aug 4 at 9:49
  • $\begingroup$ @Dedalus I'm sorry i wanted to post a new question, but got confused with this.. $\endgroup$ – Josh fisher Aug 4 at 9:52
  • $\begingroup$ No worries, just wanted to inform you. $\endgroup$ – Dedalus Aug 4 at 9:54

Say that $\alpha$ is a root of your polynomial $X^{p^n}-X-t^{-1}=0.$ Then it is obvious that if $a \in \mathbb{F}_{p^n},$ that $\alpha+a$ is a root as well, since $(\alpha+a)^{p^n}= \alpha^{p^n}+a^{p^n} = \alpha^{p^n}+a.$ So the Galois group is as claimed.

There are finite extensions with Galois groups isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$ with $n>1.$ This can be done using the Witt polynomials, see for example: Cyclic Artin-Schreier-Witt extension of order $p^2$ .

  • 2
    $\begingroup$ Everything you say is, of course, correct. But the answer is missing the piece that the given polynomial is irreducible. That follows from Artin-Schreier theory if we can show that the polynomial has no roots in $K((t))$. Mind you, I'm sure you know how to handle that, but I'm not sure all the readers are aware of all the subtleties. For example, if we replace $t^{-1}$ with $t$, then the series $u=-(t+t^{p^n}+t^{2p^n}+\cdots)\in K((t))$ is a zero of $X^{p^n}-X-t$. $\endgroup$ – Jyrki Lahtonen Aug 4 at 10:19
  • $\begingroup$ Thank you for reminding me and posting this comment. You are indeed correct about this point and it is worth pointing out. $\endgroup$ – Dedalus Aug 4 at 10:49
  • $\begingroup$ @Jyrki Lahtonen how one can show that the polynomial is irreducible? $\endgroup$ – Josh fisher Aug 10 at 17:21
  • $\begingroup$ @Joshfisher This is not hard at all. Have you used the valuation associated to your field, and the non-archimedean property of this valuation? The irreducibility follows immediately from this. $\endgroup$ – Dedalus Aug 10 at 18:14
  • $\begingroup$ @Dedalus Do you mean to use the fact that $v_{\mathfrak{P}}(X^{p^n} - X) = v_{\mathfrak{P}}(T^{-1}) = e(\mathfrak{P}/T)$ ? $\endgroup$ – Josh fisher Aug 10 at 18:20

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